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Forum: Linear Algebra Topic: Eigenvalues of Symmetric Matrices 
Demetres 
Posted: Mon Dec 03, 2018 10:12 pm


Replies: 1 Views: 55

This follows from the following characterisation of the eigenvalues of symmetric matrices. \[\lambda_k(A) = \min_{\dim(U) = nk+1} \max_{0 \neq x \in U} \frac{x^TAx}{\x\_2}\] Here the minimum is taken over all subspaces $U$ of $\mathbb{R}^n$ of dimension $nk+1$. (I chose to write $\lambda_K(A)$ r... 


Forum: Foundation Topic: Uncountably many disjoint subsets similar to \(\mathbb{R}\)? 
Demetres 
Posted: Tue Sep 26, 2017 10:09 am


Replies: 1 Views: 1234

It remained unanswered for quite some time. Since I have been asked for it, here is a solution: The answer is affirmative. Given a sequence $a = (a_1,a_2,a_3,\ldots)$, where $a_i \in \{1,2\}$ for each $i$, we construct the subset $S_a$ of $(0,1)$ as follows: Every element of $S_a$ is of the form \[ ... 


Forum: Analysis Topic: An interesting limit 
Demetres 
Posted: Thu Jul 14, 2016 10:39 am


Replies: 2 Views: 622

We have also seen this here : The basic steps are as follows: We consider independent Poisson distributions \(X_1,X_2,\ldots\) with parameter 1. We know that \(Y_n = X_1 + \cdots + X_n\) is Poisson with parameter \(n\). From the central limit theorem, \((Y_nn)/\sqrt{n}\) converges in distribution t... 


Forum: Real Analysis Topic: Inflection Point of a function 
Demetres 
Posted: Thu Jul 14, 2016 7:48 am


Replies: 2 Views: 426

I presume you can show that \(f'(1) = 1\) showing that \(x=1\) is a stationary point. Now observe that for \(0 < x < 1\) we have \(x^x < x^1 = x\). So \(x^{x^x} < x^x\), i.e. \(f(x) < 0 = f(1).\) Also, for \(x > 1\) we have \(x^x > x\) giving \(f(x) > f(1)\). So \(x=1\) is a point of inflection. 


Forum: Real Analysis Topic: Infinite Series 
Demetres 
Posted: Wed Jul 13, 2016 9:06 am


Replies: 2 Views: 420

To use Stirling's approximation observe that \[ \prod_{r=1}^n \left(1 + \frac{r}{n}\right)^{1/n} = \left( \frac{(n+1)(n+2)\cdots (n+n)}{n^n}\right)^{1/n} = \left(\frac{(2n)!}{n^n(n!)}\right)^{1/n}.\] By Stirling's approximation we have \[ \frac{(2n)!}{n^n(n!)} \sim \frac{\sqrt{4\pi n}(2n/e)^{2n}}{n^... 


Forum: Calculus Topic: floor function integral 
Demetres 
Posted: Wed Jul 13, 2016 7:47 am


Replies: 5 Views: 644

Indeed as Apostolos says, the integral diverges. What jacks's proof shows is that for every \(0 < \varepsilon < \pi/2\) we have that \(\int_{\varepsilon}^{\pi\varepsilon} \lfloor \cot{x} \rfloor \; dx = \pi + 2\varepsilon.\) However this does not mean that the value of the integral converges to\(−... 


Forum: Real Analysis Topic: Positive function 
Demetres 
Posted: Tue Jul 12, 2016 8:09 am


Replies: 1 Views: 326

If this is a Riemann integral, we use the fact that any Riemann integrable function has a point of continuity. (In fact the set of point of discontinuity has measure zero  this is Lebesgue's criterion.) So there is an interval \(I \subseteq [0,x]\) and a positive real number \(a\) such that \(f(t) ... 


Forum: Complex Analysis Topic: Analytic Function 
Demetres 
Posted: Tue Jul 12, 2016 7:36 am


Replies: 1 Views: 493

No it is not. For example \(z\) is analytic but \(\overline{z}\) is not. (As can be checked e.g. by the CauchyRiemann equations.)
By the way, "analytic" and "locally represented by a convergent power series" mean the same thing. 


Forum: Real Analysis Topic: constant function 
Demetres 
Posted: Tue Jul 12, 2016 7:26 am


Replies: 3 Views: 548

Hi Apostole,
It is a nice proof! I have not seen it anywhere. 


Forum: Real Analysis Topic: constant function 
Demetres 
Posted: Tue Jul 12, 2016 7:25 am


Replies: 3 Views: 548

Given two points \(x,y \in \mathcal{D}\) we define \[\displaystyle{D(x,y) = \frac{f(x)f(y)}{xy}}.\] Note that for \(x < y < z\) we have \[ D(x,y) = \alpha D(x,z) + (1\alpha)D(z,y)\] where \(\alpha = \alpha(x,y,z) = \displaystyle{\frac{zx}{xy} \in (0,1)}.\) It follows that either \(D(x,z) \leqsl... 


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